Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 44 / 1

Build the spherical system by adding shells from the centre outwards. The shell theorem says that a shell of mass $dM$ at radius $r$ interacts with the already assembled interior mass through potential $-GM(r)/r$. Thus its contribution is $dW=-GM(r)dM(r)/r$. Each pair is counted when its outer member is added, so no further factor of one half is needed:Integration by parts givesFor an isolated finite-mass system without a singular central point mass, the endpoint term vanishes provided the central cusp has finite binding energy. ConsequentlyThis is the spherical binding-energy mass integral. If a central singularity makes the endpoint or integral divergent, the finite-energy assumption must be checked rather than silently discarding that term.

Work in the centre-of-mass frame and define $I={\sum }_{i}m|{\mathbf{r}}_i{|}^2$, $T=\frac{1}{2}{\sum }_{i}m|{\mathbf{v}}_i{|}^2$. Differentiation givesFor each gravitational pair, the two contributions combine toSumming pairs therefore gives the exact scalar virial theoremIn a stationary equilibrium, or a bounded long-time average with vanishing average $\ddot{I}$, $2T=-W$. With $M=Nm$, the stated gravitational radius gives $W=-G{M}^2/R_g$, henceThe rms speed itself is $[GNm/R_g{]}^{1/2}$. This is a three-dimensional mean-square speed; for an isotropic system the one-component velocity dispersion is smaller by a factor of three in its square. No assumption about a uniform spatial density was used.

Take the gravitational potential to vanish at infinity. At the position of star $i$, excluding its own point-mass field, the instantaneous test-particle escape velocity obeys $v_{\mathrm{esc},i}^2=-2{\Phi }_i$. The factor of one half in the total gravitational potential energy is important:Therefore the equal-mass average isThus the rms escape speed is twice the equilibrium rms speed. This is the virial mean escape speed relation, using the same three-dimensional averaging convention as the preceding part. It concerns escape in the instantaneous system potential, rather than a subsequent collective rearrangement of the cluster.

At rest with all separations initially infinite, both initial kinetic and gravitational energies vanish, so the conserved total energy is zero. During an isolated collapse, $T+W=0$, rather than the equilibrium relation $2T+W=0$. Hence at the stated instantaneous gravitational potential energy,The rms collapse speed is $\sqrt{2}$ times the rms equilibrium speed for the same instantaneous $W$ and mass. For a perfectly radial cold infall this mean-square motion includes coherent inward streaming; it need not equal a local random velocity dispersion after subtracting that streaming. The quoted comparison follows the question's global mean-square convention.